Chemistry Calculator: Molar Mass, Molarity, pH, Stoichiometry & All Key Formulas

• The three most-calculated quantities in chemistry: molar mass (sum of atomic masses), molarity (moles per litre of solution), and pH (−log[H⁺]). Get these right and 40% of calculation problems solve themselves.
• Stoichiometry is the backbone of chemistry calculations — mole ratios from balanced equations link reactants to products. The limiting reagent determines maximum yield; always identify it first.
• pH scale: pH 7 = neutral, below 7 = acidic, above 7 = basic. A change of 1 pH unit = 10× change in [H⁺] concentration. pH + pOH = 14 at 25°C.
• For JEE: Physical chemistry (thermodynamics, equilibrium, electrochemistry) carries 30–35% weightage; Organic chemistry 30–35%; Inorganic 30–35%. Calculator-based physical chemistry problems respond best to formula mastery.

Use our free Chemistry Calculator to compute molar mass, molarity, pH, equilibrium constants, and stoichiometric quantities for any reaction.

MODULE 1: ATOMIC STRUCTURE AND MOLE CONCEPT

Molar Mass Calculator

Molar mass = Sum of (atomic mass × subscript) for each element

Common atomic masses (IUPAC, rounded for calculations):

Examples:

• H₂O: 2(1) + 16 = 18 g/mol
• H₂SO₄: 2(1) + 32 + 4(16) = 98 g/mol
• NaOH: 23 + 16 + 1 = 40 g/mol
• CaCO₃: 40 + 12 + 3(16) = 100 g/mol
• NaCl: 23 + 35.5 = 58.5 g/mol
• Fe₂O₃: 2(56) + 3(16) = 160 g/mol
• Al₂(SO₄)₃: 2(27) + 3(32 + 4×16) = 54 + 3(96) = 342 g/mol

The Mole — Key Relationships

Worked Example: How many molecules are in 9 g of water? n = 9/18 = 0.5 mol N = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules

MODULE 2: SOLUTIONS AND CONCENTRATION

Concentration Expressions

Dilution formula: M₁V₁ = M₂V₂ (moles constant on dilution)

Worked Example: Prepare 250 mL of 0.1 M NaOH. What mass of NaOH is needed? n = M × V = 0.1 × 0.250 = 0.025 mol m = n × M = 0.025 × 40 = 1.0 g NaOH

MODULE 3: ACIDS, BASES AND pH

pH and Equilibrium

pH of Common Solutions

For weak acids: [H⁺] = √(Ka × C) pH = ½(pKa − log C)

For weak bases: [OH⁻] = √(Kb × C) pOH = ½(pKb − log C), then pH = 14 − pOH

Worked Example: Find pH of 0.1 M acetic acid (Ka = 1.8 × 10⁻⁵): [H⁺] = √(1.8 × 10⁻⁵ × 0.1) = √(1.8 × 10⁻⁶) = 1.342 × 10⁻³ pH = −log(1.342 × 10⁻³) = 2.87

MODULE 4: STOICHIOMETRY

Steps for Stoichiometry Problems

1. Write and balance the equation 2. Convert given to moles (n = m/M or n = V×M for solutions) 3. Use mole ratio from balanced equation 4. Convert moles to required quantity (mass, volume, molecules)

Limiting Reagent

The limiting reagent is completely consumed first and limits the product formed.

Method:

• Calculate moles of each reactant
• Divide by stoichiometric coefficient
• Smaller value = limiting reagent

Worked Example: 10 g of H₂ reacts with 80 g of O₂. How much H₂O forms? Balanced: 2H₂ + O₂ → 2H₂O

Moles H₂ = 10/2 = 5 mol; divide by coeff (2) = 2.5 Moles O₂ = 80/32 = 2.5 mol; divide by coeff (1) = 2.5

Equal → both are limiting (exactly stoichiometric) Moles H₂O = 2 × 2.5 = 5 mol (from H₂ ratio) Mass H₂O = 5 × 18 = 90 g

Percentage Yield

% Yield = (Actual yield / Theoretical yield) × 100

MODULE 5: CHEMICAL EQUILIBRIUM

Equilibrium Expressions

For: aA + bB ⇌ cC + dD

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ (concentration equilibrium constant)

Kp = Kc(RT)^Δn (pressure equilibrium constant)

Where Δn = (moles gaseous products) − (moles gaseous reactants)

Le Chatelier's Principle Summary

MODULE 6: ELECTROCHEMISTRY

Key Electrochemical Formulas

F (Faraday constant) = 96,500 C/mol n = number of electrons transferred

MODULE 7: CHEMICAL KINETICS

Quantity	Formula	Notes
Rate law	rate = k[A]^m[B]^n	m, n from experiment
Half-life (1st order)	t½ = 0.693/k	Independent of concentration
Half-life (2nd order)	t½ = 1/(k[A]₀)	Depends on [A]₀
Integrated 1st order	ln[A] = ln[A]₀ − kt	—
Arrhenius equation	k = Ae^(−Ea/RT)	Ea = activation energy
log form	log(k₂/k₁) = Ea/2.303R × (1/T₁ − 1/T₂)	—

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MODULE 8: THERMODYNAMICS

Quantity	Formula	Unit
Enthalpy	ΔH = ΔU + ΔngRT	kJ/mol
Entropy	ΔS = q_rev/T	J/K
Gibbs free energy	ΔG = ΔH − TΔS	kJ/mol
Spontaneous reaction	ΔG < 0	—
Hess's law	ΔH_rxn = Σ ΔH_products − Σ ΔH_reactants	kJ/mol

Standard state conditions: 25°C (298 K), 1 atm, 1 M concentration

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FAQ

How do I calculate molar mass quickly?

Add atomic masses (from periodic table) × subscript for each element. For H₂SO₄: H=1×2=2, S=32×1=32, O=16×4=64; total = 98 g/mol. Keep a list of common molar masses memorised: H₂O=18, CO₂=44, NH₃=17, NaCl=58.5, CaCO₃=100, H₂SO₄=98, HCl=36.5.

What is the difference between molarity and molality?

Molarity (M) = moles of solute per litre of solution — changes with temperature (solution volume changes). Molality (m) = moles of solute per kg of solvent — temperature independent. Molality is preferred for colligative property calculations (boiling point elevation, freezing point depression). Molarity is used for most laboratory concentrations.

How do I identify the limiting reagent?

Divide the moles of each reactant by its stoichiometric coefficient in the balanced equation. The reactant with the smallest result is the limiting reagent. This is the most reliable method and works for any number of reactants.

What pH indicates a neutral solution?

pH = 7 at 25°C indicates neutral (equal [H⁺] and [OH⁻], both 10⁻⁷ M). At higher temperatures, Kw increases, so neutral pH is slightly below 7 (e.g., pH 6.13 at 60°C). For board exams and JEE: always assume pH 7 = neutral unless temperature is specified.